- #71
conway
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collinsmark said:Hello conway,
But I don't get the connection to the antenna. Even though the wave-function's expectation value may be varying with time due to the superposition of states, the atom is not radiating electromagnetic energy due to this. If it was, QM would have serious conservation of energy problems. If left completely isolated, hydrogen atoms would widdle away to nothing (assuming that it stayed in the superposition of states indefinitely [i.e same thing as saying no photons were released or absorbed]). So I think I can recognize the picture of an electron's wavefunction in a superposition of energy eigenstates in a hydrogen atom, but no, I don't see how that relates to a classical antenna.
This oscillation of charge is exactly why the atom radiates. Or absorbs, depending on the circumstance (because it might be acting as a receiving antenna). It makes a lot of sense to look at the atom this way:
1. It explains why the eigenstates are stable. They are the only states that don't have oscillating charge distributions.
2. It explains the energy transfer between the electromagnetic field and the atom. If the atom is in the ground state and it is driven at the difference frequency between the ground and the excitied states, it will oscillate at that frequency, putting it in a superposition of those two states. As an absorbing antenna, it continues to draw power from the external field and as it does, the p component grows at the expense of the s component. The more the p component grows the stronger it oscillates, and it soon reaches an equilibrium where the abosrbed energy equals the re-scattered energy.
3. It explains the decay rate (or linewidths) of the atomic spectrum. You can easily calculate the power output of the atom using the classical antenna formulas, and it gives you the correct values.
There is basically nothing that the atoms do in terms of their interaction with the electric field that you can't explain by treating them as little classical antennas.